Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Evaluate the integral $\int {{{\sin }^2}} \left( {2x + 5} \right)dx$.

Last updated date: 13th Jun 2024
Total views: 403.5k
Views today: 6.03k
Verified
403.5k+ views
Hint: To evaluate the required integral, first we will use the trigonometric formula which is given by ${\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$. After that we will use the sum rule of integration. Then, we will use the integration formula to integrate the function $\cos \left( {a\theta + b} \right)$.

In the given problem, we have to evaluate the integral $\int {{{\sin }^2}} \left( {2x + 5} \right)dx$. Let us say $I = \int {{{\sin }^2}\left( {2x + 5} \right)dx}$. Now we are going to use the trigonometric formula which is given by ${\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$. So, we can write
$I = \int {\left[ {\dfrac{{1 - \cos 2\left( {2x + 5} \right)}}{2}} \right]dx}$
In the above integral, let us use the property of integral which is given by $\int {kf\left( x \right)dx = k\int {f\left( x \right)dx} }$. So, we can write
$I = \dfrac{1}{2}\int {\left[ {1 - \cos \left( {4x + 10} \right)} \right]dx}$
In above integral, we will use the sum rule of integration which is given by $\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx = \int {f\left( x \right)dx \pm \int {g\left( x \right)dx} } }$. So, we can write
$I = \dfrac{1}{2}\left[ {\int {1\,dx - \int {\cos \left( {4x + 10} \right)dx} } } \right]$
Now we are going to use the integration formula which is given by $\int {\cos \left( {ax + b} \right)dx = \dfrac{{\sin \left( {ax + b} \right)}}{a}}$. Also note that $\int {1\,dx = x}$. After integration, we will include a constant which is called integrating constant. So, we can write
$I = \dfrac{1}{2}\left[ {x - \dfrac{{\sin \left( {4x + 10} \right)}}{4}} \right] + c$
$I = \dfrac{x}{2} - \dfrac{{\sin \left( {4x + 10} \right)}}{8} + c$
Hence, the required integral is $\dfrac{x}{2} - \dfrac{{\sin \left( {4x + 10} \right)}}{8} + c$. Hence, we can say that $\int {{{\sin }^2}} \left( {2x + 5} \right)dx = \dfrac{x}{2} - \dfrac{{\sin \left( {4x + 10} \right)}}{8} + c$ where $c$ is integrating constant.
Note: In this type of problems of integration, we must remember the trigonometric identities and formulas. Also remember the formula of integration. There is one useful trigonometric formula which is given by ${\cos ^2}\theta = \dfrac{{1 + \cos 2\theta }}{2}$.